# variable separable!

• Sep 17th 2012, 02:38 AM
lawochekel
variable separable!
$\displaystyle dy/dx=x^{-1}(2y^{\frac{1}{2}}-2y)$

have reduce to $\displaystyle \int\frac{1}{2y^{\frac{1}{2}}-2y}dy=\int\frac{1}{x}dx$

i have tried the u substitution but still unable to reduce the left-hand side to its simplest form, pls what is it am missing here?

thanks!
• Sep 17th 2012, 02:45 AM
kalyanram
Re: variable separable!
Try $\displaystyle y=u^2$ substitution.
• Sep 17th 2012, 02:54 AM
Prove It
Re: variable separable!
Quote:

Originally Posted by lawochekel
$\displaystyle dy/dx=x^{-1}(2y^{\frac{1}{2}}-2y)$

have reduce to $\displaystyle \int\frac{1}{2y^{\frac{1}{2}}-2y}dy=\int\frac{1}{x}dx$

i have tried the u substitution but still unable to reduce the left-hand side to its simplest form, pls what is it am missing here?

thanks!

\displaystyle \displaystyle \begin{align*} \int{\frac{1}{2\sqrt{y} - 2y}\,dy} &= \int{\frac{1}{2\sqrt{y}\left( 1 - \sqrt{y} \right)}\,dy} \\ &= -\int{\frac{1}{1 - \sqrt{y}}\left( -\frac{1}{2\sqrt{y}}\right) dy} \end{align*}

Now make the substitution \displaystyle \displaystyle \begin{align*} u = 1 - \sqrt{y} \implies du = -\frac{1}{2\sqrt{y}}\,dy \end{align*} and the integral becomes \displaystyle \displaystyle \begin{align*} -\int{\frac{1}{u}\,du} \end{align*}. I'm sure you can go from here :)